Решите не равенство: 5х-2 - 3-х >1
                                         3       2

5x--2/3  -- 3x/2  > 1  
10x--4  -- 9x / 6 >1 
x--4 -- 6 /6 >0
x--10 >0
x>10

[tex]\frac{5x-2}{3}-\frac{3-x}{2}>1\; |\cdot 6\\\\10x-4-9+3x>6\\\\13x>19\\\\x>\frac{19}{13}\\\\2)\quad 4sin2x-3sin(2x-\frac{\pi}{2})=5\\\\4sin2x-3sin(-(\frac{\pi}{2}-2x))=5\\\\4sin2x+3cos2x-5=0\\\\4\cdot 2sinxcosx+3(cos^2x-sin^2x)-5(sin^2x+cos^2x)=0[/tex]

[tex]8sinxcosx-8sin^2x-2cos^2x=0\; |:cos^2x\ne 0\\\\8tgx-8tg^2x-2=0\\\\t=tgx,\; 8t^2-8t+2=0\\\\4t^2-4t+1=0\\\\(2t-1)^2=0\\\\t=\frac{1}{2}\\\\tgx=\frac{1}{2}, x=arctg\frac{1}{2}+\pi n,n\in Z[/tex]